Consider the combination of $2$ capacitors $C _{1}$ and $C _{2},$ with $C _{2}> C _{1},$ when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{ C _{2}}{ C _{1}}$
JEE MAIN 2021, Diffcult
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When connected in parallel $C _{ eq }= C _{1}+ C _{2}$

When in series

$C _{ eq }^{\prime}=\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}$

$C _{1}+ C _{2}=\frac{15}{4}\left(\frac{ C _{1} C _{2}}{ C _{1}+ C _{2}}\right)$

$4\left( C _{1}+ C _{2}\right)^{2}=15 C _{1} C _{2}$

$4 C _{1}{ }^{2}+4 C _{2}^{2}-7 C _{1} C _{2}=0$

dividing by $C _{1}{ }^{2}$

$4\left(\frac{ C _{2}}{ C _{1}}\right)^{2}-\frac{7 C _{2}}{ C _{1}}+4=0$

Let $\frac{ C _{2}}{ C _{1}}= x$

$4 x^{2}-7 x+4=0$

$b ^{2}-4 ac =49-64<0$

No solution exits981-s782

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